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2 years ago

6

A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down

? Explain why

2 answers:

Gnesinka [82]2 years ago

6 0

Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

solmaris [256]2 years ago

5 0

The maximum is 1800 J since energy is not created nor destroyed, it is just transformed. In this example, gravitational potential energy is transformed to kinetic energy.

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An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The

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Answer:

$4.4443704375\times 10^{-18}\ C/m^2$

Explanation:

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$\Delta l$ = Distance charge traveled = 2 m

v = Velocity of electron = 420 m/s

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As the energy of the system is conserved we have

$q_eE\Delta l=\dfrac{1}{2}m_ev^2\\\Rightarrow E=\dfrac{1}{2}\dfrac{m_e}{q_e}\times \dfrac{v^2}{\Delta l}\\\Rightarrow E=\dfrac{1}{2}\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\times \dfrac{420^2}{2}\\\Rightarrow E=2.51094375\times 10^{-7}\ N/C$

For an infinite non conducting sheet electric field is given by

$E=\dfrac{\sigma}{2\epsilon}\\\Rightarrow \sigma=2E\epsilon\\\Rightarrow \sigma=2\times 2.51094375\times 10^{-7}\times 8.85\times 10^{-12}\\\Rightarrow \sigma=4.4443704375\times 10^{-18}\ C/m^2$

The surface charge density is $4.4443704375\times 10^{-18}\ C/m^2$

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3 years ago

A car has an initial velocity of 0m/s. It accelerates at a rate of 5m/s2 and travels 30m. What is the final velocity?

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Answer:

2.5m/s1

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2 years ago

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Answer: $a=2.42 m/s^2$

Explanation:

Given

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$m_2=1.00\times 10^{3} kg$

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$\mu =0.21$

Let T be the tension in the rope

From Diagram

$m_1g\sin \theta -T-f_r=m_1a$ -----------------1

where $f_r=friction\ force$

$f_r=\mu m_g\cos \theta$

For block $m_2$

$T=m_2a$ -----------2

From 1 & 2

$m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a$

$m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a$

$\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a$

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4 years ago

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where

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v2 is the final speed of the car

Solving for v2, we find

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DerKrebs [107]

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